Fundamentele Informatica 1 (I&E) najaar 2014 http://www.liacs.leidenuniv.nl/~vlietrvan1/fi1ie/ Rudy van Vliet rvvliet(at)liacs(dot)nl college 11, maandag 1 december 2014 6. Context-Free and Non-Context-Free Languages 6.1. The Pumping Lemma for Context-Free Languages 1
6. Context-Free and Non-Context-Free Languages 6.1. The Pumping Lemma for Context-Free Languages reg. languages FA reg. grammar reg. expression determ. cf. languages DPDA cf. languages PDA cf. grammar re. languages TM unrestr. grammar 2
A slide from lecture 3: 2.4 The Pumping Lemma Theorem 2.29. The Pumping Lemma for Regular Languages. Suppose L is a language over the alphabet Σ. If L is accepted by a finite automaton M = ( Q, Σ , q 0 , A, δ ), and if n is the number of states of M , then for every x ∈ L satisfying | x | ≥ n , there are three strings u , v , and w such that x = uvw and the following three conditions are true: 1. | uv | ≤ n . 2. | v | > 0 (i.e., v � = Λ). 3. For every i ≥ 0, the string uv i w also belongs to L . 3
A slide from lecture 3: Example 2.30. The language AnBn . Let L = { a i b i | i ≥ 0 } . 4
Now, context-free languages. Intuitively clear that PDA cannot accept AnBnCn or XX . . . 5
Now, context-free languages. Intuitively clear that PDA cannot accept AnBnCn or XX . . . Pumping lemma based on derivation in CFG (not on PDA): S ⇒ ∗ vAz ⇒ ∗ v wAy z ⇒ ∗ vw x yz S ⇒ ∗ vAz ⇒ ∗ v wAy z ⇒ ∗ vw wAy yz ⇒ ∗ vw m xy m z 6
Theorem 6.1. The Pumping Lemma for Context-Free Languages. Suppose L is a context-free language. Then there is an integer n so that for every u ∈ L with | u | ≥ n , u can be written as u = vwxyz , for some strings v , w , x , y and z satisfying 1. | wy | > 0 2. | wxy | ≤ n 3. for every m ≥ 0, vw m xy m z ∈ L Proof. . . 7
A slide from lecture 7: Definition 4.29. Chomsky Normal Form A context-free grammar is said to be in Chomsky normal form if every production is of one of these two types: A → BC (where B and C are variables) A → σ (were σ is a terminal symbol) 8
A slide from lecture 8: Theorem 4.30. (not Theorem 4.31!) For every context-free grammar G , there is another CFG G 1 in Chomsky normal form such that L ( G 1 ) = L ( G ) − { Λ } . What if Λ / ∈ L ( G ) ? 9
Number of leaf nodes in a binary tree of a given height 10
Theorem 6.1. The Pumping Lemma for Context-Free Languages. Proof Let G be CFG in Chomsky normal form with L ( G ) = L − { Λ } . Derivation tree in G is binary tree (where each parent of a leaf node has only one child). Height of a tree is number of edges in longest path from root to leaf node. At most 2 h leaf nodes in binary tree of height h : | u | ≤ 2 h . 11
At most 2 h leaf nodes in binary tree of height h : | u | ≤ 2 h . Let p be number of variables in G , let n = 2 p and let u ∈ L ( G ) with | u | ≥ n . (Internal part of) derivation tree of u in G has height at least p . Hence, longest path in (internal part of) tree contains at least p + 1 (internal) nodes. Consider final portion of longest path in derivation tree. (leaf node + p + 1 internal nodes), with ≥ 2 occurrences of a variable A . Pump up derivation tree, and hence u . 12
Pumping Lemma for CFLs ✁ ❆ ✗✔ ✁ ❆ ✁ ❆ S ✖✕ ✁ ❆ ✁ ✁ ❆ ✁ ✁ ❆ ❅ ✁ ❆ ❅ ✁ ❆ ❅ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✗✔ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ A ✁ ❆ ✖✕ ✁ ✁ ❆ ❆ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ � ✁ ✁ ❆ ❆ ✁ ❆ � ✁ ✁ ❆ ❆ ✁ � ❆ ✁ ✁ ❆ ❆ ✁ ✁ ✁ ❆ ❆ ❆ ✁ ✁ ✗✔ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ A ✁ ✁ ❆ ❆ ✁ ✖✕ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ y v w x z 13
Pumping Lemma for CFLs ✁ ❆ ✁ ❆ ✗✔ ✗✔ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ❆ S S ✖✕ ✖✕ ✁ ❆ ✁ ❆ ✁ ✁ ✁ ❆ ✁ ❆ ✁ ✁ ✁ ❆ ✁ ❆ ❅ ❅ ✁ ❆ ✁ ❆ ❅ ❅ ✁ ❆ ✁ ❆ ❅ ❅ ✁ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ✗✔ ❆ ✁ ✗✔ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ❆ ✁ ✁ A ❆ ❆ ✁ ✁ A ❆ ❆ ✁ ❆ ✁ ❆ ✖✕ ✖✕ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ❆ � ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ❆ � ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ❆ � ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ✗✔ ❆ ❆ ✁ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ✁ A ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✖✕ ✁ ❆ ✁ ❆ ✁ ✁ ❆ ❆ x ✁ ❆ ✁ ✁ ❆ ❆ ❆ ✁ ❆ ✁ ✁ ✁ ❆ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ❆ ✁ ❆ ✁ ✁ ✁ ❆ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ � ✁ ❆ ✁ ✁ ✁ ❆ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ � ✁ ❆ ✁ ✁ ✁ ❆ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ � ✁ ❆ ✁ ✁ ✁ ❆ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ✁ ✗✔ ❆ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ✁ ❆ ❆ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ✁ ✁ A ❆ ❆ ✁ ❆ ✖✕ ✁ ✁ ❆ ❆ y v z v w z ✁ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ❆ ✁ ✁ ❆ ❆ ✁ ✁ ✁ ❆ ❆ ❆ y w x 14
Application of pumping lemma: mainly to prove that a language L cannot be generated by a context-free grammar. How? Find a string u ∈ L with | u | ≥ n that cannot be pumped up! What is n ? What should u be? What can v , w , x , y and z be? 15
Suppose that there exists context-free grammar G with L ( G ) = L . Let n be the integer from the pumping lemma. Pumping lemma: For every u ∈ L with | u | ≥ n , there are five strings v , w , x , y and z such that u = vwxyz and the following three conditions are true: 1. | wy | > 0 2. | wxy | ≤ n 3. for every m ≥ 0, vw m xy m z ∈ L 16
Suppose that there exists context-free grammar G with L ( G ) = L . Let n be the integer from the pumping lemma. We prove: NOT ( For every u ∈ L with | u | ≥ n , there are five strings v , w , x , y and z such that u = vwxyz and the following three conditions are true: 1. | wy | > 0 2. | wxy | ≤ n 3. for every m ≥ 0, vw m xy m z ∈ L ) 17
Suppose that there exists context-free grammar G with L ( G ) = L . Let n be the integer from the pumping lemma. We prove: There exists u ∈ L with | u | ≥ n , such that NOT ( there are five strings v , w , x , y and z such that u = vwxyz and the following three conditions are true: 1. | wy | > 0 2. | wxy | ≤ n 3. for every m ≥ 0, vw m xy m z ∈ L ) 18
Suppose that there exists context-free grammar G with L ( G ) = L . Let n be the integer from the pumping lemma. We prove: There exists u ∈ L with | u | ≥ n , such that for every five strings v , w , x , y and z such that u = vwxyz NOT ( the following three conditions are true: 1. | wy | > 0 2. | wxy | ≤ n 3. for every m ≥ 0, vw m xy m z ∈ L ) 19
Suppose that there exists context-free grammar G with L ( G ) = L . Let n be the integer from the pumping lemma. We prove: There exists u ∈ L with | u | ≥ n , such that for every five strings v , w , x , y and z such that u = vwxyz NOT all of the following three conditions are true: 1. | wy | > 0 2. | wxy | ≤ n 3. for every m ≥ 0, vw m xy m z ∈ L ) 20
Suppose that there exists context-free grammar G with L ( G ) = L . Let n be the integer from the pumping lemma. We prove: There exists u ∈ L with | u | ≥ n , such that for every five strings v , w , x , y and z such that u = vwxyz if 1. | wy | > 0 2. | wxy | ≤ n then NOT ( 3. for every m ≥ 0, vw m xy m z ∈ L ) 21
Suppose that there exists context-free grammar G with L ( G ) = L . Let n be the integer from the pumping lemma. We prove: There exists u ∈ L with | u | ≥ n , such that for every five strings v , w , x , y and z such that u = vwxyz if 1. | wy | > 0 2. | wxy | ≤ n then 3. there exists m ≥ 0, such that vw m xy m z does not belong to L 22
Example 6.3. Applying the Pumping Lemma to AnBnCn AnBnCn = { a i b i c i | i ≥ 0 } Choose u = . . . 23
Example 6.3. Applying the Pumping Lemma to AnBnCn AnBnCn = { a i b i c i | i ≥ 0 } Choose u = a n b n c n 24
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